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The problem states that a current \( I \) passes through a resistor with resistance \( R \), and according to Ohm's Law, the voltage drop across the resistor is given by:

\[
V = IR
\]

We are asked to use differentials to show that the relative error in calculating \( I \) is approximately the same (in magnitude) as the relative error in \( R \), given that \( V \) is constant.

### Step-by-Step Solution:

1. **Start with Ohm’s Law**:  
   From \( V = IR \), we solve for the current \( I \):
   \[
   I = \frac{V}{R}
   \]

2. **Take the differential of the equation**:  
   Differentiating \( I = \frac{V}{R} \) with respect to \( R \), we get:
   \[
   dI = -\frac{V}{R^2} dR
   \]
   Since \( V \) is constant, its differential \( dV = 0 \).

3. **Relative error in \( I \)**:  
   Now, divide the equation by \( I = \frac{V}{R} \) to get the relative error in \( I \):
   \[
   \frac{dI}{I} = -\frac{1}{R} dR
   \]

4. **Relative error in \( R \)**:  
   The relative error in \( R \) is:
   \[
   \frac{dR}{R}
   \]

5. **Compare the relative errors**:  
   From \( \frac{dI}{I} = -\frac{dR}{R} \), we observe that the relative error in \( I \) is the same in magnitude as the relative error in \( R \), since \( \frac{dI}{I} \) is equal to \( -\frac{dR}{R} \).

Thus, the relative error in calculating \( I \) is approximately the same in magnitude as the relative error in \( R \).

---

Would you like further clarification on any steps? Here are 5 additional related questions to explore the topic:

1. What happens to the current \( I \) if the resistance \( R \) increases while the voltage \( V \) remains constant?
2. How do differentials help in approximating relative errors in other physical systems?
3. Could we extend this analysis to power dissipation \( P = I^2 R \) in the resistor?
4. How would the analysis change if the voltage \( V \) were not constant but subject to a small error?
5. What is the importance of using differentials in engineering applications involving error propagation?

**Tip:** Always remember that relative errors give a proportional sense of how much uncertainty affects a measurement, which is critical in engineering and physics applications.

If a current I passes through a resistor with resistance R, Ohm's Law states that the voltage drop is V = IR. If V is constant and R is measured with a certain error, use differentials to show that the relative error in calculating I is approximately the same (in magnitude) as the relative error in R.

Explore how to use differentials to show that the relative error in calculating current I is approximately the same as the relative error in resistance R, using Ohm's Law with constant voltage V. Learn step-by-step through differentiation and error propagation techniques.

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